Real Analysis

Ordinals

Ordinal Numbers

Suppose there are sets \(X\), \(Y\), and orders on them, namely \((X, <)\) and \((Y, <)\). Say they have the same order-type if there exists a bijection \(f : X \rightarrow Y\) such that \(x < y\) if and only if \(f(x) < f(y)\). \(f\) is called an order-isomorphism.

Recall \(X\) is well-ordered if every non-empty subset of \(X\) has a least element.

Observations

If \(\alpha\) is an ordinal then \(S(\alpha) = \alpha \cup \{\alpha\}\) is an ordinal, called the successor ordinal.

If \(\alpha\) is an ordinal and \(\beta \in \alpha\) then \(\beta\) is an ordinal.

If \(A\) is a set of ordinals then \(\sup A = \bigcup A\) is an ordinal.

Theorems

Theorem. Any well-ordered set is order-isomorphic to some ordinal \(\alpha\).

\(\omega\), called the limit ordinal is the first infinite ordinal. It is not the successor of any ordinal (i.e. \(\omega \neq S(\alpha)\) for any ordinal \(\alpha\)). \(\epsilon_0\) is the first ordinal such that \(\epsilon_0 = \omega^{\epsilon_0}\). \(\omega_1\) is the first uncountable ordinal.

The principle of transfinite induction: Suppose \(J\) is well-ordered. A set \(A \subset J\) is inductive implies \(A = J\).

Applications

Theorem. There exists a set \(K\) in \(\mathbb{R}^2\) that intersect every line in the plane twice.

Proof (depends on the axiom of choice). Let \(L\) be the set of all lines in \(\mathbb{R}^2\) (so \(|L| = |\mathbb{R}|\)). Well-order \(L\) with the type \(J\) of the first ordinal with the same cardinality as \(\mathbb{R}\) (all elements of \(J\) have cardinality less than that of \(\mathbb{R}\)). Write \(L = \{L_\alpha\}_{\alpha \in J}\). Let \[\begin{align}
A = \{\alpha \in J : \exists \text{ set } K_\alpha \text{ such that } & (1) |K_\alpha| < |\mathbb{R}| \text{ and}\\
& (2) \text{ no three points are colinear} \text{ and} \\
& (3) |K_\alpha \cap L_\beta| = 2 \text{ if } \alpha \leq \beta \text{ and} \\
& (4) K_\beta \subset K_\alpha \text{ if } \beta < \alpha\}
\end{align}\] We need to show that \(A\) is inductive, so \(A = J\).

Base case: Clearly, \(1 \in A\) (let \(K_1\) be the set of two points on \(L_1\)).

Inductive step: If \(S_\alpha \subset A\), let \(K = \bigcup\limits_{\beta < \alpha} K_\beta\). \(K\) has the same cardinality as \(\mathbb{R}\) (by (1)) and has no three point colinear (by (2) and (3)). The set of all lines through \(K_\alpha\) have cardinality less than that of \(\mathbb{R}\) so it can’t hit all of \(L_\alpha\). Pick 1 or 2 points to form \(K_\alpha = K \cup \{\text{1 or 2 extra points}\}\). So \(\alpha \in A\). QED.